Let $\omega_1,\omega_2$ be the circles with center $O_1,O_2$ respectively. Let the angle bisector of $\angle O_1AO_2$ be $l$. Let $l\cap\omega_1=C(\ne A)$ and $l\cap\omega_2=D(\ne A)$. Let $O_3$ be the center of $\odot(BCD)$.
Notice that $O_1O_3$ and $O_2O_3$ are the perpendicular bisectors of $CD$ and $BD$ respectively. So, $$\angle O_3O_1O_2=\angle O_3O_1C+\angle CO_1O_2=\frac{\angle BO_1C}{2}+\angle O_2O_1B-\angle BO_1C=\frac{\angle AO_1C}{2}.$$Similarly, $$\angle O_3O_2O_1=\angle BO_2O_3+\angle BO_2O_1=\frac{\angle AO_2D}{2}.$$As $CD$ is bisector of $\angle O_1AO_2$ we have $$\angle AO_2D=180^\circ-2\angle O_2AD=180^\circ-\angle O_1AC=\angle AO_1C .$$Thus, $O_3O_1=O_3O_2$ as desired. $\blacksquare$